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I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$ Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$...

How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?

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    $\begingroup$ I have not read Spivak's book, but I would have guessed $\lambda$ is the constant map $x\rightarrow \lambda x$, which does not influence $f$'s differentiablity since $\lambda f$ is just multiplying $f$ by a constant. $\endgroup$ – Bombyx mori Sep 1 '12 at 20:35
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$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$ is a bijection and $\lambda$ and $\lambda^{-1}$ are both continuously differentiable. Note that $\lambda'(z) = \lambda$ for all $z \in \mathbb{R}$.

Let $g = \lambda^{-1}\circ f$. Suppose the theorem is true for $g$. Then there is an open set $V'$ containing $a$ and an open set $W'$ containing $g(a)$ such that $g:V'\to W'$ has a continuous inverse $g^{-1}:W'\to V'$ which is differentiable and for all $y\in W'$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$ Then $\lambda(W')$ is open and $f = \lambda\circ g:V'\to \lambda(W')$ has a continuous inverse $g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$.

By the chain rule, for all $z \in \lambda(W')$, $(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$

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  • $\begingroup$ Why is $\lambda$ a bijection? $\endgroup$ – mathemather May 27 '18 at 12:07
  • $\begingroup$ $\lambda$ is non-singular (since $\det f'(a) \not= 0$). Hence it estabilish a one-to-one correspondence between the function domain and image (i.e. it is a bijection) $\endgroup$ – Antonio Horta Ribeiro Mar 28 at 7:53
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Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.

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  • $\begingroup$ I dont understand the last statement. How does $g \circ \lambda ^{-1} $being continuously differentiable imply the theorem is true for $f$? $\endgroup$ – mathemather May 27 '18 at 12:11
  • $\begingroup$ Notice how he placed the parentheses around $g\circ\lambda^{-1}$. This has no effect on computing the compositions; he simply did it to show you that $g\circ\lambda^{-1}$ is in fact the inverse of $f$ that we are looking for! And in addition, this inverse is continuously differentiable since it is just the map $g$ and we generated $g$ as a result of applying the Inverse function Theorem to $\lambda^{-1} \circ f(a)$, so $g$ certainly satisfies all the properties that we'd want from our inverse. $\endgroup$ – john fowles Jun 7 '18 at 0:01

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